题目内容

如图,已知AC平分∠PAQ,点B、B′分别在边AP、AQ上,如果添加一个条件,即可推出AB=A B′,那么该条件不可以是(    )

A.BB′⊥AC     B.CB=CB′     C.∠ACB=∠ACB′    D.∠ABC=∠AB′C
B.

试题分析:如图:∵AC平分∠PAQ,点B,B′分别在边AP,AQ上,

A:若BB′⊥AC,在△ABC与△AB′C中,∠BAC=∠B′AC,AC=AC,∠ACB=∠ACB′,∴△ABC≌△AB′C,∴AB=AB′;
B:若BC=B′C,不能证明△ABC≌△AB′C,即不能证明AB=AB′;

C:若∠ACB=∠ACB′,则在△ABC与△AB'C中,∠BAC=∠B′AC,AC=AC,∴△ABC≌△AB′C,∴AB=AB′;
D:若∠ABC=∠AB′C,则∠ACB=∠ACB′∠BAC=∠B′AC,AC=AC,∴△ABC≌△AB′C,∴AB=AB′.
故选B.
练习册系列答案
相关题目

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网